A gaseous mixture containing equal mole sof `H_(2),O_(2)` and He is subjected to series of effusion steps. The composition (by moles) of effused mixture after 4 effusion steps is `x:1:y` rspectively. Then find the value of `((x)/(y)).`

Explanation

To solve the problem, we need to analyze the effusion of the gaseous mixture containing equal moles of \( H_2 \), \( O_2 \), and \( He \). We will use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Identify Molar Masses**:
- Molar mass of \( H_2 \) = 2 g/mol
- Molar mass of \( O_2 \) = 32 g/mol
- Molar mass of \( He \) = 4 g/mol

2. **Calculate the Rate of Effusion**:
According to Graham's law, the rate of effusion \( R \) is given by:
\[ R \propto \frac{1}{\sqrt{M}} \]
where \( M \) is the molar mass.

3. **Calculate Ratios for Each Gas**:
- For \( H_2 \) and \( O_2 \):
\[ \frac{R_{H_2}}{R_{O_2}} = \frac{\sqrt{M_{O_2}}}{\sqrt{M_{H_2}}} = \frac{\sqrt{32}}{\sqrt{2}} = \frac{4}{1} \quad \text{(Ratio of rates)} \]
- For \( O_2 \) and \( He \):
\[ \frac{R_{O_2}}{R_{He}} = \frac{\sqrt{M_{He}}}{\sqrt{M_{O_2}}} = \frac{\sqrt{4}}{\sqrt{32}} = \frac{2}{4} = \frac{1}{2} \quad \text{(Ratio of rates)} \]

4. **Apply Effusion Steps**:
Since we are considering 4 effusion steps, we raise the ratios to the power of the number of steps:
- For \( H_2 \) to \( O_2 \):
\[ \left(\frac{R_{H_2}}{R_{O_2}}\right)^4 = \left(4\right)^4 = 256 \]
- For \( O_2 \) to \( He \):
\[ \left(\frac{R_{O_2}}{R_{He}}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

5. **Determine the Composition After Effusion**:
Let the initial moles of \( H_2 \), \( O_2 \), and \( He \) be equal, say \( n \) moles each. After 4 effusion steps, the remaining moles will be:
- Moles of \( H_2 \) = \( n \times 256 \)
- Moles of \( O_2 \) = \( n \)
- Moles of \( He \) = \( n \times \frac{1}{16} \)

6. **Express the Final Composition**:
The final composition by moles can be expressed as:
\[ H_2 : O_2 : He = 256n : n : \frac{n}{16} \]
Simplifying this gives:
\[ 256 : 1 : \frac{1}{16} \]

7. **Find the Value of \( \frac{x}{y} \)**:
From the composition \( 256 : 1 : \frac{1}{16} \), we set:
- \( x = 256 \)
- \( y = \frac{1}{16} \)

Therefore:
\[ \frac{x}{y} = \frac{256}{\frac{1}{16}} = 256 \times 16 = 4096 \]

### Final Answer:
The value of \( \frac{x}{y} \) is \( 4096 \).